
theorem
  for p be odd Prime holds p divides ((p + 1) choose ((p + 1)/2))
  proof
    let p be odd Prime;
    reconsider k = (p - 1)/2 as Nat;
    p + p > p + 1 by XREAL_1:6,INT_2:def 4; then
    (2*p)/2 > (p + 1)/2 by XREAL_1:74; then
    p divides (p choose ((p + 1)/2)) by NEWTON02:119; then
    A1: p divides ((2*k + 1) choose k) by NP0;
    ((2*k + 1) choose k) divides 2*((2*k + 1) choose k); then
    p divides (2*((2*k + 1) choose k)) by A1,INT_2:9; then
    p divides ((2*(k + 1) choose (k + 1))) by NP1;
    hence thesis;
  end;
